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I have a real-life problem which essentialy boils down to the following: Given n biased coins, given to you ordered from highest to low bias, determine the last coin (by index) whose bias is still above some cutoff, say 0.5.

If possible I would also like to be able to say, given the number of flips I've made for each coin, what the probability is for a given coin that it is the last coin above the cutoff.

I know I can use pearson-clopper intervals to determine for each coin seperately at some desired level of significance whether or not it is above the cutoff, but this fails to take into account the extra information that the biases are in decreasing order.

My knowledge of statistics is quite minimal so what I'm really looking for is a push into the right direction (Bayesian statistics? ).

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  • $\begingroup$ What else do you know about the coins? For example: there is no solution if all the coins have the same bias, or each has bias greater then cutoff. $\endgroup$
    – Tim
    May 19, 2016 at 8:47
  • $\begingroup$ @Tim It is possible that they all have a bias greater, or smaller than the cutoff, in this case the 'solution' is that there is no coin which satisfies the property. You can also assume no coin has bias equal to another coin. $\endgroup$
    – Arnaud
    May 19, 2016 at 9:03

1 Answer 1

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Interesting problem. Here is how I would set it up.

For coin $i$, let $\theta_i$ represent the success probability, $y_i$ the number of successful flips, and $n_i$ the number of trials. Then assume

$$ y_i \stackrel{ind}{\sim} Bin(n_i,\theta_i). $$

Let $\theta=(\theta_1,\ldots,\theta_n)$. Then to incorporate the ordering you mention, I would use the prior

$$ p(\theta) \propto \mathrm{I}(\theta_1>\theta_2>\cdots>\theta_n). $$

(Any prior that incorporates this ordering could be used.)

Let $y=(y_1,\ldots,y_n)$, the posterior is

$$ p(\theta|y) \propto \left[ \prod_{i=1}^n \theta_i^{y_i}(1-\theta_i)^{n_i-y_i} \right] \mathrm{I}(\theta_1>\theta_2>\cdots>\theta_n) $$

and, if nothing else, you can draw samples from this posterior using MCMC.

Then your question of interest is to calculate probabilities of this nature

$$ P(\theta_i>c,\theta_{i+1}<c|y) $$

where $c$ is your cutoff. This can be estimated for $i=1,\ldots,n-1$ and augmented with $P(\theta_1<c|y)$ and $P(\theta_n>c|y)$ which are the only other possibilities.

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  • $\begingroup$ Thank you, seems like the way to go! Just one small question: would it be useful to use the posterior as the new prior each time I flip a coin? Or is there little value in updating the prior each measurement? $\endgroup$
    – Arnaud
    May 19, 2016 at 11:34
  • $\begingroup$ If you are using MCMC to obtain samples from the posterior, then it will be a huge waste to rerun it for every flip. If you really do want a sequential analysis, e.g. an updated posterior after every flip, then you probably want to think harder about how to approximate those posteriors. $\endgroup$
    – jaradniemi
    May 19, 2016 at 17:48

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